Let me teach you a secret: You don't need to memorize any trig identities, you just need to know Euler's formula and one or two more facts:

    exp(ix) = cos(x) + i*sin(x)
    [cos(x)]^2 + [sin(x)]^2 = 1
    i*i = -1

That is, the complex number exp(ix), which lies on the unit circle in the complex plane, has real component cos(x) and imaginary component sin(x). Consequently, the line from 0 to exp(ix) in the complex plane makes an angle of x radians with the real axis.

Any complex number can be written as an exponent like rexp(ix), where x is the angle and r is the distance from the origin in the complex plane. Using this understanding, we can interpret complex number multiplication as a rotation in the complex plane:

    (a * exp(ix)) * (b * exp(iy)) = ab * exp(i (x+y))

Ok, the above was a digression, let's get to the trig identities. Let's derive the double angle formula. We can start with the point on the unit circle representing an angle of 2x, and use Euler's formula to simplify it one way:

    exp(i*2x) = cos(2x) + i * sin(2x)

Ok, we can also use the fact that a^(bc) = (a^b)^c to write the same expression a different way, then FOIL it:

    exp(i*2x) = [exp(i*x)]^2
    = [cos(x) + i*sin(x)]^2
    = [cos(x)]^2 + (i*i)*[sin(x)]^2 + 2i * cos(x) * sin(x)
    = [cos(x)]^2 - [sin(x)]^2 + 2i * cos(x) * sin(x)

Now, comparing the two expressions, we see that

    cos(2x) + i * sin(2x) = [cos(x)]^2 - [sin(x)]^2 + 2i * cos(x) * sin(x)

Taking real and imaginary parts of both sides, we find that:

    cos(2x) = [cos(x)]^2 - [sin(x)]^2
    sin(2x) = 2 * cos(x) * sin(x)

This works for most identities you want to derive, and there is a variation for things like cosh and sinh. This trick came in handy on calc 3 exams whenever I couldn't remember the silly trig identities that were required to perform integrals.